detail sap design

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0 Abstract: This project is a structural analysis and design of a new building for Nablus Municipality. The building is composed of a reinforced concrete structure with a steel dome and steel arches. The approach is to do a preliminary design using 1D and 2D models. The loads considered are gravity loads for the reinforced concrete parts and gravity and wind loads for the steel structure parts. A 3D analysis using SAP2000 is done and elements are designed accordingly. The 3D analysis matches approximately the 1D analysis in most parts of the project. Complete design is done for all structural parts, and few structural design details are provided.

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  • 0

    Abstract:

    This project is a structural analysis and design of a new building for

    Nablus Municipality. The building is composed of a reinforced concrete

    structure with a steel dome and steel arches.

    The approach is to do a preliminary design using 1D and 2D models.

    The loads considered are gravity loads for the reinforced concrete parts and

    gravity and wind loads for the steel structure parts. A 3D analysis using

    SAP2000 is done and elements are designed accordingly.

    The 3D analysis matches approximately the 1D analysis in most parts

    of the project. Complete design is done for all structural parts, and few

    structural design details are provided.

  • 1

    Chapter1: Introduction

    1.1 Project Description:

    The project is structural analysis and design of a proposed new

    building for Nablus Municipality. First a preliminary analysis and design using

    1D and 2D models are made, then the analysis and design are done using 3D

    model using SAP2000 program.

    The building is composed of 3 stories above the ground, and an

    underground parking. The area of each story is about 1580m2, and its height

    is 4m.

    The project has 2 structural joints which divide the building into three

    parts: A,B, and C from left to right. It is composed of both reinforced concrete

    and steel parts. The large dome over part B is made of steel and the vault

    over the entrance is made of steel arches. The rest of the structure which

    includes the two domes over the stair roofs over part B, and the dome over

    minaret are made of reinforced concrete. All these descriptions are illustrated

    on fig.1.1.

    Fig. 1.1 3D view of municipality building

    The architectural plans were designed in year 2007 by Haythem

    Alsaadi, who is an architectural student at AN-Najah National University.

  • 2

    1.2 Design Codes:

    In this project, the American Concrete Institute (ACI) Code2005 is used in

    designing concrete parts, and the British Code (BS 5950) in designing steel parts.

    1.3 Materials

    1. Concrete

    Concrete strength for all concrete parts is B350f'c=280 Kg/cm

    Modulus of elasticity equals 2.5*105 Kg/cm2.

    Unit weight is 2.5 ton/m3.

    2. Steel:

    Modulus of elasticity equals 2.04*106 Kg/cm2.

    Steel yield strength:

    For steel reinforcement, is 4200 kg/cm2.

    For rolled steel, is 2750 kg/cm2.

    3. The unit weights of the main materials used are shown in table1.1:

    Table1.1. Density of the main materials used

    Density(ton/m3) Material

    2.5 Reinforced concrete

    1.2 Bricks

    1.5 Filler

    2.7 Masonry

    2.5 Tiles

    2.3 Mortar

    2.3 plastering

    1.9 Selected filler (compacted

    base coarse)

    0.04 Polycarbonate

  • 3

    1.4 Loads:

    1. Dead Loads:

    Dead loads are composed of own weight of the slabs, beams, columns, walls,

    Domes, and superimposed dead loads from the partitions and tiles.

    o The super imposed dead loads are composed of:

    Bricks, mortar and filling as shown below

    Thus their weight is equal to

    0.03*2.5+2.3*0.02+0.1*1.5 =0.27ton/m

    partitions: consist of bricks 10cm thickness and plastering 1.5cm

    from each side.

    Using false ceiling height 0.45m under concrete slab, the partitions height will

    equal 4-0.45-0.25=3.3m.

    Average distance between partitions= 6m

    (0.1*1,2*3.4+.03*2.3*3.3)/6 =0.1 ton/m

    Total super imposed load on slab =0.27+0.1 =0.37ton/m2.

    2. Live loads:

    For the municipality building the live loads will be taken as 400 kg/m2.

    3. Wind load

    The Assessment of wind load according to British Standard should be made

    as follows :

    1. The basic wind speed V appropriate to the district where the structure is to be erected

    2. The basic wind speed is multiplied by factors S1, S2 , S3 to give the design wind speed Vs for the part under consideration

    Vs = V * S1 *S2 * S3

    3. The design wind speed is converted to dynamic pressure q using the relationships

    q = k Vs2

  • 4

    4. The dynamic pressure is then multiplied by an appropriate pressure coefficient Cp to give the pressure p exerted at any point on the surface of a building

    p = Cp*q Cp = Cpe -Cpi

    If the value of the pressure coefficient Cp is negative this indicates that p is a

    suction as distinct from a positive pressure

    In Nablus the average basic wind speed is 80km/hr, at 10m above

    ground in an open situation.

    S1 is a topography factor, it will be taken 1 for level terrain.

    S3 is a factor that takes into account of the degree of security and the

    period of time in years during which there will be exposure to wind ,

    normally wind loads on complete structure and buildings should be

    calculated at S3 = 1 with the following exceptions :

    o Temporary structures. o Structures where a shorter period of exposure to the wind

    may be expected o Structures where a longer period of exposure to the wind

    may be required o Structure where greater than normal safety is required

    Since our building is not one of those exception then S3 = 1

    k = 0.613 in SI units (N/m2 and m/s) " British code page 145".

    S2 is a factor which accounts for ground roughness, building size and

    height above ground. This factor is taken from Table1.2.

  • 5

    Table1.2. Ground Roughness, building size and height above ground, factor S2

    H

    (m)

    (1)open country with no

    obstructions

    (2) open country with

    scattered windbreaks

    (3)country with many

    windbreaks , small

    town , outskirts of

    large cities

    (4) surface with

    large and frequent

    obstructions ,city

    centers

    Class*

    Class Class Class

    A B C A B C A B C A B C

    3 or

    less 0.83 0.78 0.73 0.72 0.67 0.63 0.64 0.6 0.55 0.56 0.52 0.47

    5 0.88 0.83 0.78 0.79 0.74 0.7 0.7 0.65 0.6 0.6 0.55 0.5

    10 1 0.95 0.9 0.93 0.88 0.83 0.78 0.74 0.69 0.67 0.62 0.58

    15 1.03 0.99 0.94 1 0.95 0.91 0.88 0.83 0.78 0.74 0.69 0.64

    20 1.06 1.01 0.96 1.03 0.98 0.94 0.95 0.9 0.85 0.79 0.75 0.7

    30 1.09 1.05 1 1.07 1.03 0.98 1.01 0.97 0.92 0.9 0.85 0.79

    40 1.12 1.08 1.03 1.1 1.06 1.01 1.05 1.01 0.96 0.97 0.93 0.89

    50 1.14 1.1 1.06 1.12 1.08 1.04 1.08 1.04 1 1.02 0.98 0.94

    60 1.15 1.12 1.08 1.14 1.1 1.06 1.1 1.06 1.02 1.05 1.02 0.98

    80 1.18 1.15 1.11 1.17 1.13 1.09 1.13 1.1 1.06 1.1 1.07 1.03

    100 1.2 1.17 1.13 1.19 1.16 1.12 1.16 1.12 1.09 1.13 1.1 1.07

    120 1.22 1.19 1.15 1.21 1.18 1.14 1.18 1.15 1.11 1.15 1.13 1.1

    140 1.24 1.2 1.17 1.22 1.19 1.16 1.2 1.17 1.13 1.17 1.15 1.12

    160 1.25 1.22 1.19 1.24 1.21 1.18 1.21 1.18 1.15 1.19 1.17 1.14

    180 1.26 1.23 1.2 1.25 1.22 1.19 1.23 1.2 1.17 1.2 1.19 1.16

    200 1.27 1.24 1.21 1.26 1.24 1.21 1.24 1.21 1.18 1.22 1.21 1.18

    * Class A. All units of cladding, glazing and roofing and their immediate fixings

    and individual members of unclad structures, class B. All buildings and

    structures where neither the greatest horizontal dimension nor the greatest

    vertical dimension exceeds 50m,

    This building will be considered case 4, classC.

    The internal pressure coefficient "Cpi" will be taken -0.3, corresponding

    to case that all faces are impermeable.

  • 6

    The external pressure coefficient " Cpe" will be taken from the table 1.3.

    Table 1.3 Pressure coefficients Cpe for pitch roofs of rectangular clad buildings

    Pressure coefficients Cpe for pitch roofs

    building

    height* ratio roof angle

    wind angle =0

    degree

    wind angle =90

    degree

    Wind

    ward

    face

    Leeward

    face

    Wind

    ward

    face

    Leeward

    face

    h/w 0.5 0 -0.8 -0.4 -0.8 -0.4

    5 -0.9 -0.4 -0.8 -0.4

    10 -1.2 -0.4 -0.8 -0.6

    20 -0.4 -0.4 -0.7 -0.6

    30 0 -0.4 -0.7 -0.6

    45 0.3 -0.5 -0.7 -0.6

    60 0.7 -0.6 -0.7 -0.6

    0.5 < h/w < 1.5 0 -0.8 -0.6 -1 -0.6

    5 -0.9 -0.6 -0.9 -0.6

    10 -1.1 -0.6 -0.8 -0.6

    20 -0.7 -0.5 -0.8 -0.6

    30 -0.2 -0.5 -0.8 -0.8

    45 0.2 -0.5 -0.8 -0.8

    60 0.6 -0.5 -0.8 -0.8

    1.5 < h/w < b 0 -0.7 -0.6 -0.9 -0.7

    5 -0.7 -0.6 -0.8 -0.8

  • 7

    10 -0.7 -0.6 -0.8 -0.8

    20 -0.8 -0.6 -0.8 -0.8

    30 -1 -0.5 -0.8 -0.7

    40 -0.2 -0.5 -0.8 -0.7

    50 0.2 -0.5 -0.8 -0.7

    60 0.5 -0.5 -0.8 -0.7

    * h is building height, and w is building length

    Wind angle will be taken zero. Building height ratio (h/w)

  • 8

    Table 1.4. Wind pressure on the steel dome

    point

    no.

    elevation

    (m) S2

    angle of

    inclination

    Cpe

    (win ward

    face)

    Pressure

    (N/m)

    at the wind

    ward face

    Cpe

    (leeward

    face)

    Pressure

    (N/m)

    at the leeward

    face

    1 17.32 0.668 70 0.7 135 -0.6 -66.1

    2 19 0.688 60 0.7 143.3 -0.6 -43

    3 20.25 0.702 40 0.2 74.57 -0.433 -19.8

    4 20.91 0.708 10 1.2 -163.54 -0.4 -15.2

    For Steel Arches:

    The arches will be divided into 5 areas marked by 5 points, as shown on

    Fig1.2. Table 1.5 gives wind pressure values calculated in a similar way as

    given previously.

    Fig.1.3. Steel Arches

  • 9

    Table 1.5. Wind pressure on the steel arches.

    point

    no.

    Elevation

    (m) S2

    Angle of

    Inclination

    Cpe at

    windward

    face

    pressure

    (N/m) at

    windward

    face

    Cpe at

    leeward

    face

    pressure

    (N/m) at

    leeward

    face

    1 12 0.645 50 0.433 92.29 -0.533 -29.34

    2 13 0.668 37.6 0.355 88.46 -0.45 -20.26

    3 13.75 0.688 25 -0.2 14.33 -0.4 -14.33

    4 14.21 0.702 18.06 -0.555 -38.03 -0.4 -14.91

    5 14.37 0.708 0 -0.8 -75.86 -0.4 -15.17

    1.5 Loads combinations:

    For concrete structures:

    Ultimate load=1.2*DL (Dead load)+1.6*LL(live load).

    For steel structures:

    1. Ultimate load=1.4*DL+1.6*LL

    2. Ultimate load=1.0*DL+1.4*WL(wind load)

    3. Ultimate load=1.2*(DL+LL+WL)

  • 10

    Chapter2: Preliminary design:

    This Chapter provides manual analysis and design.

    2.1 Structural Systems

    As stated before, this project is divided into 3 parts( part A, B, and C

    from left to right) by 2 structural joints as shown on Fig. 2.1.

    The steel dome over part B is composed of hollow section steel

    members covered by Polycarbonate, and the cylindrical roof is composed of

    3-hinged steel I-section arches covered by purlins and polycarbonate.

    All floors, except roof #3, are one way ribbed slabs on main drop

    beams carried by columns, as shown on figures 2.1 through 2.6. Roof #3 is

    designed as one way solid slab on main dropped beams on columns as

    shown on Fig. 2.6.

    Basement walls are used around the parking floor, and all the exterior

    walls are composed of concrete, masonry, and blocks.

  • 11

    2.2 Concrete Domes Analysis and Design :

    There are 3 concrete domes, one on the minaret and two on the stair roofs.

    These are the major equations of the domes :

    For dead load

    Meridian stress (C)

    =wR(K1)/d+W(K3)/R.(a)

    Hoop stress() =-wR(k2)/d-

    W(K3)/Rd ..(b)

    Where,

    w: uniformly distributed load per

    meter square on the area

    W: concentrated load

    R: radius of the dome

    d: thickness of the dome

    K1,2,3: constants which depend on the semi central angle and the angle at

    which the stress is measured, are given in Table 2.1

    For live and snow loads:

    Meridian stress(C) =-q*R/2d.(c)

    Hoop stress () =-q*R*(cos 2c)/2d.(d)

    Where,

    q: uniformly distributed load per square meter on horizontal projection.

    c: the angle measured from the vertical to the point at which the stress is

    to be measured.

  • 12

    Table 2.1. K Factors

    c k1 k2 k3

    0 0.5 0.5

    10 0.505 -0.48 5.3

    20 0.516 -0.425 1.37

    30 0.537 -0.33 0.64

    40 0.566 -0.2 0.38

    50 0.608 -0.034 0.27

    51.48 0.618 0 0.26

    60 0.667 0.167 0.21

    70 0.747 0.402 0.18

    80 0.838 0.68 0.16

    90 1 1 0.16

    1. Minaret Dome:

    Concrete dome with ring beam below.

    Dimensions:

    Radius (R)=3 m, Rise = 2.22 m, c = 75

    Thickness(d)= 10 cm (range 7.5cm

    15cm).

    Loads :

    Dead load(w) =own weight=1*1*0.1*2.5=0.25 t/m

    Live and snow loads(q) =0.15 t/m (on projected area).

    Super imposed load, concentrated(W) =0.05 t.

    Angle of super imposed load = 10 degree.

    Applying Eqs. A to d, the stresses on the dome are given in Table 2.2.

  • 13

    Table2.2. Hoop and Meridian stresses for minaret dome

    c C dead (ton/m2)

    dead (ton/m2)

    C live and snow

    (ton/m2)

    live and snow

    (ton/m2)

    C total (ton/m2)

    total (ton/m2)

    0 4.50 -4.50 -2.25 -2.25 2.25 -6.75

    10 4.55 4.32 -2.25 -0.92 2.30 3.40

    20 4.10 2.96 -2.25 1.50 1.85 4.46

    30 4.13 2.37 -2.25 2.14 1.88 4.51

    40 4.31 1.44 -2.25 0.25 2.06 1.69

    50 4.61 0.21 -2.25 -1.94 2.36 -1.73

    51.48 4.68 -0.04 -2.25 1.70 2.43 1.66

    60 5.04 -1.29 -2.25 -1.83 2.79 -3.12

    70 5.63 -3.05 -2.25 0.45 3.38 -2.60

    75 6.59 -4.58 -2.25 1.32 4.34 -3.26

    All tension and compression stresses are much less than concrete

    capacity, where concrete compression capacity is 2800 ton/m2, and the

    tension concrete capacity is approximately equal to 280 ton/m2.

    Thus only shrinkage steel is needed for both directions(meridian and

    hoop).

    As =0.2%*10*100 =2 cm2 use 4 8mm/m.

    2. Domes on the stair roofs:

    Concrete dome with ring beam below.

    Dimensions:

    Radius =1.8 m, Rise =0.85 m, c = 58

    Thickness 10 cm (range 7.5cm 15cm).

    Loads :

    Dead load = only its own weight (w)=1*1*0.1*2.5=0.25 t/m.

    Live and snow loads (q) =0.15 t/m (on the projected area).

    Applying Eqs. A to d, the stresses on the dome are given in table 2.2.

  • 14

    Table2.3. Meridian and Hoop Stresses for domes over stair roofs

    c C dead (ton/m2)

    dead (ton/m2)

    C live and snow

    (ton/m2)

    live and snow

    (ton/m2)

    C total (ton/m2)

    total (ton/m2)

    0 2.25 -2.25 -1.35 -1.35 -1.35 -1.35

    10 2.27 2.16 -1.35 -0.55 0.92 1.61

    20 2.32 1.91 -1.35 0.90 0.97 2.81

    30 2.42 1.49 -1.35 1.29 1.07 2.77

    40 2.55 0.90 -1.35 0.15 1.20 1.05

    50 2.74 0.15 -1.35 -1.16 1.39 -1.01

    51.48 2.78 0.00 -1.35 1.02 1.43 1.02

    58 3.00 -0.75 -1.35 -1.10 1.65 -1.85

    All tension and compression stresses are much less than concrete

    capacity, where concrete compression capacity is 2800 ton/m2, and the

    tension concrete capacity is approximately equal to 280 ton/m2.

    Thus only shrinkage steel is needed for both directions(meridian and

    hoop).

    As =0.2%*10*100 =2 cm2 use 4 8mm/m.

  • 15

    2.3 Slabs Analysis and Design

    Ribbed Slabs Analysis and Design:

    Referring to Fig. 2.2 the thickness of slab, Hmin,,. is given using ACI Table 2.4.

    Table 2.4: minimum thickness of one way solid slabs and beams

    Member

    Minimum thickness, h

    One end

    continuous

    Two ends

    continuous

    simply

    supported Cantilever

    One way

    solid slab 24

    ln

    28

    ln

    20

    ln

    10

    ln

    Beams 5.18

    ln

    21

    ln

    16

    ln

    8

    ln

    Hmin.= larger of ( 485/21= 23.1 cm, 360/18.5=19.5cm, 500/21=23.8 cm)=

    23.8cm 25cm.

    Figure below shows a proposed section of the ribbed slab:

    Thus the own weight is calculated as:

    Own weight= (0.08*.52+0.12*0.17)*2.5+0.4*0.17*1.2=0.2366 t/m/rib=

    0.455t/m2.

    Wu= 1.2*(0.455+0.27)+1.6*0.4= 1.51 t/m2=0.79t/m/rib.

  • 16

    Slab1 (S1) in parts A and C :

    Slab 1 load (in ton/m)

    Slab 1 reaction diagram (in ton)

    Slab 1 Shear force diagram (in ton)

    Slab 1 bending moment diagram (in ton.m)

    Max. shear at distance d =20 cm from the face of column= 1.91-

    0.79*(0.15+0.2)=1.633 ton.

    (Assumed columns dimensions are 30*70 cm.)

    Nominal shear=1.633/0.75=2.178 ton

    Vc= 0.53* cf *b*d*1.1=0.53* 280 *12*20*1.1*10-3=2.34ton>Vu no need for

    shear reinforcement.

    Max. neg. moment= 1.34 ton.m/rib

    0079.280*20*12

    34.1*10*61.211(*

    4200

    280*85.02

    5

    0033.014

    .min yF

  • 17

    As= db** = 9.120*12*0079.0 cm22 rib/12 top steel.

    Min. neg. moment = 1.19 ton.m/rib

    007.280*20*12

    19.1*10*61.211(*

    4200

    280*85.02

    5

    As =1.675 cm2 2 12 /rib

    So use 212 top steel for all spans except the exterior ends of the exterior

    spans

    Max. +ve moment= 0.99 ton.m/rib

    0012.0)280*20*52

    99.0*10*61.211(*

    4200

    280*85.02

    5

    As= 0.00127*52*20= 1.32cm2

    As min.= 8.020*12*0033.0**.min dbw cm2

  • 18

    Slab 1 Shear force diagram (in ton)

    Slab 1 bending moment diagram (in ton.m)

    Max. shear at distance d =20 cm from the face of column= 1.93-

    0.79*(0.2+0.15)=1.653 ton.

    (Assumed columns dimensions to be 30*70 cm.)

    Nominal shear=1.653/0.75=2.2 ton

    Vc= 0.53* cf *b*d*1.1=0.53* 280 *12*20*1.1*10-3=2.34ton>Vu no need for

    shear reinforcement.

    Max. neg. moment= 1.34 ton.m/rib

    0079.280*20*12

    34.1*10*61.211(*

    4200

    280*85.02

    5

    0033.014

    .min yF

  • 19

    0013.0)280*20*52

    02.1*10*61.211(*

    4200

    280*85.02

    5

    As= 0.0013*52*20= 1.36cm2

    As min.= 8.020*12*0033.0**.min dbw cm2

  • 20

    Slab3 bending moment diagram

    Max. shear at distance d =20 cm from the face of column= 2.32-

    0.79(0.2+0.15)=2 ton

    (Assumed columns dimensions to be 30*70 cm.)

    Ultimate shear=2/0.75=2.67 ton

    Vc= 0.53* cf *b*d*1.1=0.53* 280 *12*20*1.1*10-3=2.34ton10cm use 1 8mm stirrup/10 cm.

    If Vn< Vc/2 ,no need for shear reinforcement.

    Vc*0.75/2=.88 ton = 2.32-0.79*X1X1=1.82m

    So at distance 0 to 1.82m (measured from left end of span 4) use 1 8mm

    stirrup/10 cm, otherwise no need for shear reinforcement.

  • 21

    Max. neg. moment at span 4= 1.73 ton.m /rib

    01.280*20*12

    73.1*10*61.211(*

    4200

    280*85.02

    5

    0033.014

    .min yF

  • 22

    Max. +ve moment at span 1 and 2 = 0.71 ton.m/rib

    0009.0)280*20*52

    71.0*10*61.211(*

    4200

    280*85.02

    5

    As= 0.0009*52*20= 0.945cm2

    As min.= 8.020*12*0033.0**.min dbw cm2< 0.945cm2 .

    Use rib/102 bottom steel.

    For span 3 use min area of steel =.8 cm , 210 / rib.

    Check if the compression depth smaller than flange thickness:

    a =bf

    fA

    c

    YS

    **85.0

    *

    =

    52*280*85.0

    4200*26.2=0.77cm

  • 23

    Slab Shear force diagram (ton/m)

    Slab bending moment diagram (ton.m)

    Max. shear at distance d =20 cm from the face of column= 1.42-

    0.79*(0.2+0.15)= 1.143 ton

    (Assumed columns dimensions to be 30*70 cm.)

    Ultimate shear=1.243/0.75=1.524 ton

    Vc= 0.53* cf *b*d*1.1=0.53* 280 *12*20*1.1*10-3=2.34ton>Vu no need for

    shear reinforcement.

    Max. neg. moment= 1.05 ton.m/rib

    0061.280*20*12

    05.1*10*61.211(*

    4200

    280*85.02

    5

    0033.014

    .min yF

  • 24

    00028.0)280*20*52

    22.0*10*61.211(*

    4200

    280*85.02

    5

    As= 0.00028*52*20 =0.29 cm

    As min.= 8.020*12*0033.0**.min dbw cm2>0.29 cm

    Use rib/102 bottom steel, for all spans.

    Check if the compression depth smaller than flange thickness:

    a =bf

    fA

    c

    YS

    **85.0

    *

    =

    52*280*85.0

    4200*57.1=0.53cm

  • 25

    Beam Load Diagram (in ton/m)

    Beam Reactions Diagram (in ton)

    Beam Shear Force Diagram (in ton)

    Beam Bending Moment Diagram (in ton.m)

    Flexure Design:

    +ve moment on span 1 =13.28 ton.m

    0061.0)280*45*30

    28.13*10*61.211(*

    4200

    280*85.02

    5

    >0.0033

    As=0.0061*45*30=8.23cm24 18

    +ve moment on span 3 =10.94 ton.m

    00497.0)280*45*30

    94.10*10*61.211(*

    4200

    280*85.02

    5

    >0.0033

    As=0.00497*45*30=6.7cm23 18

  • 26

    Span 2 will be reinforced with min. bottom steel reinforcement

    As min.=0.0033* 45*30 =4.45 cm use 2 18

    -ve. Moment on span1 = 13.53 ton.m

    0062.0)280*45*30

    53.13*10*61.211(*

    4200

    280*85.02

    5

    >0.0033

    As=0.0062*30*45=8.4 cm24 18

    -ve. Moment on span2 = 10.75 ton.m

    00488.0)280*45*30

    75.10*10*61.211(*

    4200

    280*85.02

    5

    >0.0033

    As=0.00488*30*45=6.59 cm23 18

    Shear Design

    Span1:

    Shear Force at distance d=45 cm from the face of the right support =

    19.45- 7.05*(0.15+0.45)=15.22 ton.

    Vn= uV =

    75.0

    22.15=20.29 ton

    Vc= 0.53* cf *b*d= Vc= 0.53* 280 *30*45= 11.97 ton.

    VS=Vn-VC= 20.29-11.97=8.32 ton

    dF

    V

    S

    A

    y

    SV

    *

    VS

  • 27

    S

    AV =0.044, Using 8mm stirrups AV=1cm2

    S=044.0

    1=22.7cm=22.5cm ok. use 1 8mm stirrup/20 cm c/c.

    Shear Force at distance d=45 cm from the face of the left support of

    span1 = 13.69- 7.05*(0.15+0.45)=9.46 ton

    Vn= uV =

    75.0

    46.9=12.61 ton.

    Vc= 0.53* cf *b*d= Vc= 0.53* 280 *30*45= 11.97 ton.

    VS=Vn-VC= 12.61-11.97=0.64ton

    dF

    V

    S

    A

    y

    SV

    *

    VS0.0034use min. shear

    reinforcement.

    max. spacing =min. of(d/2, 60 cm)= min. of(22.5, 60 cm)=22.5cm

    S

    AV =0.025, Using 8mm stirrups AV=1cm2

    S=025.0

    1=40cm>22.5cm ok. use 1 8mm stirrup/20 cm c/c.

    If Vn< Vc/2 ,no need for shear reinforcement.

    Vc*0.75/2=4.5 ton = 13.69-7.05*X1X1=1.3m

    -4.5=13.69-7.05*X2X2=2.6m

  • 28

    So at distance 1.3m to 2.6 m (measured from left end) use 1 8mm stirrup/40

    cm.

    Span2:

    Use min. shear reinforcement at the ends(1 8mm stirrup/20 cm), and at

    distance 1.3m to 2.6m measured from left end use 1 8mm stirrup/40 cm.

    Span3:

    Use min. shear reinforcement at the ends(1 8mm stirrup/20 cm), and at

    distance 1.85m to 3.1m measured from left end use 1 8mm stirrup/40 cm.

    Beam 3 in part B in the ground floor (BM3) :

    Hmin.= cm9.4621

    301015

    H= 70cm, D= 65cm.

    Width= 30cm.

    Ultimate load on all spans =

    mton /96.63.0*)25.07.0(*5.2*2.152.0

    41.3

    Slab 2 Load Diagram (in ton/m)

    Slab1 Reaction Diagram (in ton)

  • 29

    Beam Load Diagram (in ton/m)

    Beam Reaction Diagram (in ton)

    Beam Shear Force Diagram (in ton)

    Beam Bending Moment Diagram (in ton.m)

    Flexure Design:

    +ve moment on span1,5 =21.11 ton.m

    0046.0)280*65*30

    11.21*10*61.211(*

    4200

    280*85.02

    5

    >0.0033

    As=0.0046*65*30=8.93cm24 18

    +ve moment on span 2,4 =1.88 ton.m

    0003.0)280*65*30

    88.1*10*61.211(*

    4200

    280*85.02

    5

  • 30

    Asmin=0.0033*65*30=6.43cm23 18

    +ve moment on span 3 =40 ton.m

    00905.0)280*65*30

    40*10*61.211(*

    4200

    280*85.02

    5

    0.0033

    As=0.0043*30*65=8.44 cm24 18

    -ve. Moment on span2,4 = 49.46 ton.m

    011.0)280*65*30

    46.49*10*61.211(*

    4200

    280*85.02

    5

    >0.0033

    As=0.011*30*65=21.45 cm29 18

    Shear Design

    Max. shear at distance d=65 cm from the face of the support = 35.33-

    6.96*(0.15+0.65)=29.76 ton.

    Vn= uV =

    75.0

    76.29= 39.68ton

    Vc= 0.53* cf *b*d= Vc= 0.53* 280 *30*65= 17.29 ton.

    VS=VN-VC= 29.76-17.29=12.47 ton

    dF

    V

    S

    A

    y

    SV

    *

    VS

  • 31

    025.0)024.0,025.0(.max)( .min ofS

    AV

  • 32

    Load assigned on slab 1

    Slab 1 reaction diagram

    Beam F1:

    Hmin.= cm8.645.18

    2.02.12

    H= 70cm, D= 65cm.

    Width= 30cm.

    Ultimate load= mton /23.43.0*)25.07.0(*5.2*2.152.0

    99.1

    Frame loads diagram

  • 33

    Axial force diagram

    Shear force diagram

  • 34

    Bending moment diagram

    a. Beam flexure design:

    +ve moment=49.82 ton.m

    0116.0)280*65*30

    82.49*10*61.211(*

    4200

    280*85.0

    2

    5

    >0.0033 ok.

    As=0.0116*30*65=22.62 cm28 20

    -ve moment on the left end=15.98ton.m

    0034.0)280*65*30

    98.15*10*61.211(*

    4200

    280*85.0

    2

    5

    >0.0033ok.

    As=0.0034*30*65=6.7 cm24 16

    -ve moment on the right end=42.72ton.m.

    0097.0)280*65*30

    72.42*10*61.211(*

    4200

    280*85.0

    2

    5

    >0.0033ok.

    As=0.0097*30*65=19 cm210 16

    b. Beam shear design:

  • 35

    Shear force at d distance from face of left column= 23.62-

    4.23*(0.2+0.65)=19.6ton.

    Vn= uV =

    75.0

    6.19=26.13 ton

    Vc= 0.53* cf

    *b*d= Vc= 0.53* 280 *30*65*10-3= 17.3 ton.

    VS=VN-VC= 26.13-17.3=8.83 ton

    dF

    V

    S

    A

    y

    SV

    *

    VS

  • 36

    Shear force at d distance from face of right column= 27.99-

    4.23*(0.2+0.65)=24.4ton.

    Vn= uV =

    75.0

    4.24=32.53 ton.

    Vc= 0.53* cf

    *b*d= Vc= 0.53* 280 *30*65*10-3= 17.3 ton.

    VS=VN-VC= 32.53-17.3=15.23 ton

    dF

    V

    S

    A

    y

    SV

    *

    VS

  • 37

    All the secondary beams are hidden, H=25cm, W=30cm.

    Loads=own weight+ partitions weight

    Loads=1.2*0.25*0.3*2.5+0.64=0.865ton/m.

    Beam load diagram

    Beam shear force diagram

    Beam bending moment diagram

    Flexure Design:

    Max +ve moment on span4 =1.08 ton.m

    00243.0)280*20*30

    08.1*10*61.211(*

    4200

    280*85.0

    2

    5

  • 38

    -ve. Moment on span1 = 1.49 ton.m

    0034.0)280*20*30

    49.1*10*61.211(*

    4200

    280*85.0

    2

    5

    >0.0033

    As=0.0034*30*20=2.04 cm23 10

    Use 3 10 top steel for all spans.

    Use minimum bottom and top steel(3 10) for all the secondary beams.

    Shear Design

    Max. shear at distance d=20 cm from the face of the support = 2.09-

    0.87*(0.15+0.2)=1.8 ton.

    Vn= uV =

    75.0

    8.1= 2.4ton

    Vc= 0.53* cf *b*d= Vc= 0.53* 280 *30*20= 5.32 ton.

    Vn< Vc/2=2.66 ton no need for shear reinforcement.

  • 39

    2.5 Columns analysis and design:

    B3 (in ground floor) Reaction Diagram

    Design the worst column (has the max. axial load=62.06ton)

    Assume column dimensions: 30cm*70cm.

    Max. load on column= 62.06*4+1.2(0.3*0.65*4*2.5*4)=257.6ton

    Check if r

    lk u*